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This question is related to Question 2 of my previous posting.

Question. Let $\mu$ be a Radon measure on a compact Hausdorff space $\Omega$ and $L^{\infty}(\Omega,\mu)$ the set of essentially bounded Borel measurable functions on $\Omega$. Suppose that $S$ is the set of bounded lower or upper semi-continuous functions on $\Omega$. Does $S$ generate $L^{\infty}(\Omega,\mu)$ as a $C^*$-algebra?

It suffices to consider whether the indicator function of any Borel set is obtained from $S$ by algebraic operations and (essential-supremum) norm limits. If necessary, you may assume $\Omega$ to be second countable. Thank you.

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No. The bounded Baire class one functions on $[0,1]$ are stable under uniform limits and hence constitute a C*-algebra. This C*-algebra contains every semicontinuous function on [0,1]. Every function in $L^\infty[0,1]$ is equal almost everywhere to a Baire class two function, but not a Baire class one function. (The previous version of my answer neglected this essential point.)

See http://www.encyclopediaofmath.org/index.php/Baire_classes

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  • $\begingroup$ Thank you very much Nik for your clear answer and the reference! $\endgroup$ – Masayoshi Kaneda Aug 18 '12 at 10:28
  • $\begingroup$ Sure. General rule of thumb --- ignore my first answer, wait for the correction ... $\endgroup$ – Nik Weaver Aug 18 '12 at 14:58
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Any upper or lower semicontinuous function is continuous almost everywhere in the sense of Baire category (since it is a pointwise limit of a sequence of continuous functions, at least when $\Omega$ is compact metrizable). The algebra of Baire-a.e. continuous functions is itself a $C^\ast$-algebra. So the answer to your question is no, once we show that there exists a function $f \in L^\infty$ that is not $\mu$-equivalent to a Baire-a.e. continuous function. For an example we may take any indicator function of a set $S$ such that both $\mathrm{supp}\ \mu\restriction S$ and $\mathrm{supp}\ \mu\restriction (\Omega \setminus S)$ equal $\Omega$.

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